question 2\nthe wavelength (λ) and frequency (ν) of light are related through the equation: c =…

question 2\nthe wavelength (λ) and frequency (ν) of light are related through the equation: c = λ×ν\nwhere:\nc = speed of light (3.00×10⁸ m·s⁻¹)\nλ = wavelength (m)\nν = frequency (s⁻¹)\nusing the following emission spectrum:\ncalculate the frequency for each of the 7 emission lines (1 nm = 1×10⁻⁹m):\na) violet (450 nm)\nb) indigo (470 nm)\nc) blue (490 nm)\nd) green (520 nm)\ne) yellow (620 nm)\nf) orange (630 nm)\ng) red (690 nm)

question 2\nthe wavelength (λ) and frequency (ν) of light are related through the equation: c = λ×ν\nwhere:\nc = speed of light (3.00×10⁸ m·s⁻¹)\nλ = wavelength (m)\nν = frequency (s⁻¹)\nusing the following emission spectrum:\ncalculate the frequency for each of the 7 emission lines (1 nm = 1×10⁻⁹m):\na) violet (450 nm)\nb) indigo (470 nm)\nc) blue (490 nm)\nd) green (520 nm)\ne) yellow (620 nm)\nf) orange (630 nm)\ng) red (690 nm)

Answer

Explanation:

Step1: Rearrange the formula for frequency

Given $c = \lambda\times\nu$, we can solve for $\nu$ as $\nu=\frac{c}{\lambda}$.

Step2: Convert wavelength to meters

For each wavelength value in nanometers, convert it to meters by multiplying by $10^{-9}$.

Step3: Calculate frequency for each line

a) Violet

$\lambda = 450\ nm=450\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{450\times 10^{-9}\ m}\approx6.67\times 10^{14}\ s^{-1}$

b) Indigo

$\lambda = 470\ nm = 470\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{470\times 10^{-9}\ m}\approx6.38\times 10^{14}\ s^{-1}$

c) Blue

$\lambda = 490\ nm=490\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{490\times 10^{-9}\ m}\approx6.12\times 10^{14}\ s^{-1}$

d) Green

$\lambda = 520\ nm = 520\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{520\times 10^{-9}\ m}\approx5.77\times 10^{14}\ s^{-1}$

e) Yellow

$\lambda = 620\ nm=620\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{620\times 10^{-9}\ m}\approx4.84\times 10^{14}\ s^{-1}$

f) Orange

$\lambda = 630\ nm=630\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{630\times 10^{-9}\ m}\approx4.76\times 10^{14}\ s^{-1}$

g) Red

$\lambda = 690\ nm=690\times 10^{-9}\ m$ $\nu=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m\cdot s^{-1}}{690\times 10^{-9}\ m}\approx4.35\times 10^{14}\ s^{-1}$

Answer:

a) $6.67\times 10^{14}\ s^{-1}$ b) $6.38\times 10^{14}\ s^{-1}$ c) $6.12\times 10^{14}\ s^{-1}$ d) $5.77\times 10^{14}\ s^{-1}$ e) $4.84\times 10^{14}\ s^{-1}$ f) $4.76\times 10^{14}\ s^{-1}$ g) $4.35\times 10^{14}\ s^{-1}$